Class 12 Notes

I. Basics of Logarithms
   A. Definition
   B. Graphs
   C. Multiplication and exponentiation properties
   D. The Natural Logarithm
II. Logarithms and Equations
    A. Exponential equations
    B. Change of base
    C. Logarithmic equations


I. Basics of Logarithms

A. Definition

Our last class was about exponentials, and this class is about the
inverse of the exponential: the logarithm.

The idea of a logarithm is simple. We know that

2^4 = 16

We define the logarithm in this case by

log_2 16 = 4

The rule is very simple: if

b^x = a

then

log_b a = x

Thus a logarithm is simply another way of writing an exponential. As
with exponentials, we call b the base. The quantity a is called the
argument.

Evaluating logarithms is simple, one you understand what they mean. If
we write

log_2 16,

we mean "the power to which I have to raise 2 in order to come up with
16." In this case the answer is 4, since 2^4 = 16.

Practice for the class: evaluate log_5 25, log_5 (5^2), log_3 81, log_x x^3

Notice that the logarithm doesn't have to be an integer. For example

4^(3/2) = 4 . 4^(1/2) = 4 . 2 = 8

Thus we can also write

log_4 8 = 3/2

More practice: evaluate log_4 32, log_9 27

Finally, notice that logarithms of negative arguments are undefined;
the argument is always positive. That's because

log_2 (-4)

would mean "what power should I raise 2 to in order to get
-4?" That question has no answer. There is no number to which I can
raise 2 and come up with -4, or any other negative number, because any
power of a positive number is still positive.

Similarly, the argument can't be zero. There's no power to which I can
raise 2 and come up with zero.

Finally, note that the most common logarithm base is 10, so to save
time we don't write a number for the base for base 10. In other words,
if you have just

log x

without a base written, the base is assumed to be 10.


B. Graphs

Since a logarithm is just the inverse of an exponetial, graphs of
logarithms look just like graphs of exponentials turned on their
sides.

Graph: log_2 x

Notice that, just as exponential functions are always positive (unless
they are multiplied by a negative number), logarithmic graphs have the
corresponding property that they only exist when x is positive. Their
graphs are entirely to the right of x = 0.


C. Multiplication and exponentiation properties

Logarithms have some properties involving multiplying or
exponentiating them, just like exponents do.

Recall the rule for multiplying exponents:

b^x . b^y = b^(x + y)

For example

2^3 . 2^4 = 2^7

Now notice what this implies for logarithms. We know that 

log_2 (2^3) = 3
log_2 (2^4) = 4
log_2 (2^7) = 7

Thus

log_2 (2^3) + log_2 (2^4) = log_2 (2^7) = log_2 (2^3 . 2^4)

This is a general rule:

log_b (b^x) = x
log_b (b^y) = y
log_b (b^(x+y)) = x+y,

so

log_b (b^x) + log_b (b^y) = log_b (b^(x+y)) = log_b (b^x . b^y)

But if this is true for any b, x, and y, it must be true for all
positive numbers (since logarithms only make sense for positive
numbers). Thus we have a general rule:

log_b (x . y) = log_b x + log_b y

We can use the exact same reasoning for division. Since

b^x / b^y = b^(x-y)

we know that

log_b (x / y) = log_b x - log_b y.

Finally, we can use this argument for exponentiation. We know that

(b^x)^y = b^(x.y)

Thus

log_b (b^(x.y)) = log_b ((b^x)^y) = y log_b (b^x)

But this is true for any positive b, so we must have a general rule:

log_b (x^y) = y log_b x

for any x, y, and b.


D. The Natural Logarithm

Just like the number e comes up all the time in applications where
something changes continually, its logarithm comes up all the time
too. We often take log_e.

Because this occurs so often, we have a special notation for it:

log_e x = ln x

It's just a shorter way of writing log_e. The n in ln stands for
"natural", because log_e is called the natural logarithm.

II. Logarithms and Equations

A. Exponential equations

One of the main reasons to introduce logarithms is to give us a tool
to solve equations involving exponentials.

The idea is simple: explot the properties of logarithms to turn
exponential equations into algebraic ones. Here's a simple example:

2^x = 8

The answer is pretty obviously 3. To solve this with a logarithm, we
can take log_2 of both sides:

log_2 (2^x) = log_2 8
x = 3

In this case the logarithm was fairly pointless. It's more useful if
the answer isn't so obvious. For example

2^(2x^2 + x) = 8

Taking log_2 of both sides:

2x^2 + x = 3
2x^2 + x - 3 = 0
(2x + 3)(x - 1) = 0
x = -3/2, 1

Similarly, logarithms are useful if the answer isn't a whole number or
a whole fraction, for example:

2^x = 14
log_2 (2^x) = log_2 14
x = log_2 14

In this case we wrote the answer down in terms of a logarithm. Of
course this is of limited use, since we need a calculator to evaluate
it.


B. Change of base

Notice that we could have taken a logarithm to a different base and
gotten similar results:

2^x = 14
log (2^x) = log 14
x log 2 = log 14
x = log 14 / log 2

Why is this useful? Well, for one reason, we've written the answer in
terms of just logarithms with a base of 10. That means we can solve
any problem like this just by knowing the values of log base
10. That's handy, because calculators have a button for that.

It also points out something important. We found x = log_2 14 and x =
log 14 / log 2, which means that

log_2 14 = log 14 / log 2

But the choice of 2 and 14, and a base of 10, was arbitrary. I could
have put any numbers there and gotten the same answer. Thus we have
another general rule:

log_b a = log_c a / log_c b

This is called the change of base formula, because it lets us write
the logarithm using one base in terms of the logarithm using a
different base.


C. Logarithmic equations

We can also solve equations that contain logarithms in them by
exponentiating. For example:

log_3 (2x + 1) = 2

To solve this, we take 3^ on both sides:

3^[log_3 (2x + 1)] = 3^2
2x + 1 = 9
2x = 8
x = 4

The rule here is simple: get the logarithm by itself, then take both
sides to the base that matches the base of the logarithm.

If there is more than one logarithm, combine them into a single one
first using the addition / subtaction rules. For example:

log (7x - 5) - log (x - 5) = 1

First we combine the logarithmic terms:

log [(7x - 5) / (x - 5)] = 1

Now we raise to a power:

10^{log [(7x-5) / (x-5)]} = 10^1
(7x-5) / (x-5) = 10

Now we solve the resulting equation:

7x - 5 = 10x - 50
-3x = -45
x = 15

Practice: solve log_2 x + log_2 (x - 2) = 3